I saw the paper. It said CRR is appropriate for lognormal distribution. I wonder if I want to build the binomial tree of normal distribution, may I do that??

yes, you can still use a binomial tree, but you should have each point being the previous point plus or minus a factor, rather than each point multiplied by or divided by a factor.

QuoteOriginally posted by: gjlipmanyes, you can still use a binomial tree, but you should have each point being the previous point plus or minus a factor, rather than each point multiplied by or divided by a factor.For example, like dS=mu dt +sigma dz, there is two directions. One is up, and the other is down.For up, S(t+delta t)=S(t)+mu*delta t+sigma*(delta)^(0.5).For down, S(t+delta t)=S(t)+mu*delta t-sigma*(delta)^(0.5).Is it correct?

Not exactly,You can say that for an up/down move the stock price evolves according to the binomial approximation:S+u with a probability p.S+d with a probability (1-p).You need to work out u ,d and p by matching first and second moments, etc. Like in the Lognormal case, you need to introduce an additional relationship to have a square system of equations and unknowns... I dunno if there are any popular ones that have been derived, I that think gjlipman was suggesting you set u=d... I can't see a better one to use.Regards,Sam

Here is way of getting different |u| and |d| but same up and down probability. Letting X~p G(a,b) mean X is Gaussian under prob law p with mean a and st dev b,dS will be BM with drift mu and diffusion coef sigma, if for infesm. small time incrmt. hdS ~p G(mu*h,sigma*sqrt(h)), or as lim h goes to 0dS=mu*dt+sigma*dB. Discretise this sodS=mu*h+sigma*sqrt(h). (h~iid sqrt(h)*G(0,1))We can get rid of one parameter mu by letting it equal the r-q (riskless rate minus div yield) and adjusting the prob law p (Girsanov's theorem)or we can leave mu and fix p, letting up prob and down prob be equal to 1/2. Then the tree will be given byS+u=S+mu*h+sigma*sqrt(h) with prob 1/2 during time h 1S+d=S-mu*h-sigma*sqrt(h) with prob 1/2 during time h 2 matching moments, we must have: E(dS)=mu*h (defn of BM w/ drift) 3 Var(dS)=(sigma^2)*h (ditto ) 4multiply 1 by 0.5 and 2 by 0.5 and add to get (3) E(dS) = 0.5(mu*h+mu*h)=mu*h (mean is prob wgted sum of outcomes)dS will differ from mu*h or mean of dS by amounts +/-sigma*sqrt(h) by 1 and 2, multiply the squares of these differences by their probs 1/2 and add them to getVar(dS)= 0.5*{dS-E(dS)}^2+0.5*{dS-E(dS)} =0.5*{(mu*h+sigma*sqrt(h)-mu*h)^2}+0.5*{(mu*h-sigma*sqrt(h)-mu*h)^2} =0.5*(sigma*sqrt(h))^2+0.5*(sigma*sqrt(h))^2 =(sigma^2)*h so the two moments match as wanted.If you want to let mu equal r-q from the beginning, you can discretise the BM given by dS~q G[(r-q)*h,sigma*sqrt(h)] instead, but you must either handwave about eqln retns being equal to carry cost, or assume riskless hedge poss so drift equals riskless rtn less divds over h, or invoke Girsanov if asked whence it came.Stripping the fancy measure theory the key to Girsanov is to note that Bm is defined in terms of prob law, which will always have a mean and std dev as it is Gaussian. So if you want to remove the drift, just shift the prob density function so G becomes G*=GD, where D is a discount factor. BM ~p G(a,b) ~q G*(0,b), but q is obtained from p multiplied by factor D to compensate for missing drift. A waiter is carrying a tray on a ship which is sinking, but cannot bend his knees, so he tilts the tray slightly to adjust for the ship's listing. Hope this isn't too unclear, I am rotten at explaining things.

Sorry typo in lequaltion two:should be S+dS=S+mu*h-sigma*sqrt(h), and not S+dS=S-mu*h-sigma*sqrt(h)there is a + sign in front of mu and not -, so only the sigma term changes

Hi,I am also working on a binomial tree to compute option premiums and greeks for options whose underlying follows normal distribution.I tried in both ways (either requiring that the probability of an up movement is the same as the probabiltiy of a down movement, or by imposing that the up and down movements are the same), and I decided to check the results comparing the premiums with the premiums obatined using the CEV model when the alpha parameter is zero.With my surprise, things don't seem to work very well.... I wonder if any of you is getting similar results to mine.I am using as values: European Call Option, Expiry: 1year, Stock Price: 50,Continuous return on the asset: 0Risk free rate: 0 (just to get the same result whether I am using tree with pu=pd or u=d)Num of iteractions: 1000I am trying with strikes: 30, 40, 50, 60 and 70 and CEV returns me the following premiums for the call:Volatility | K = 30 | K=40 | K=50 | K=60 | K=70----------------------------------------------------------------------------- 15 | 20.6359 |12.2668 | 5.9842 | 2.2668 | 0.6360 30 | 24.4981 |17.6156 | 11.9649 | 7.6262 | 4.5334 45 | 29.0176 |23.0118 | 17.7460 | 13.2869 | 9.6437 60 | 32.7105 |27.5098 | 22.7469 | 18.4810 | 14.7447 75 | 35.4834 |30.9786 | 26.7410 | 22.8157 | 19.2351 90 | 37.5634 |33.6278 | 29.8635 | 26.3029 | 22.9722100 | 38.6687 |35.0504 | 31.5627 | 28.2316 | 25.0792while the binomial normal tree returns me the following ones:dS | K = 30 | K=40 | K=50 | K=60 | K=70-----------------------------------------------------------------------------0.4743 | 20.6353 | 12.2678 | 5.9826 | 2.2678 | 0.63530.9487 | 24.5356 | 17.6286 | 11.9653 | 7.6286 | 4.53561.4230 | 29.6934 | 23.3982 | 17.9479 | 13.3982 | 9.69341.8974 | 35.2572 | 29.2732 | 23.9306 | 19.2732 | 15.25722.3717 | 40.9806 | 35.1845 | 29.9132 | 25.1845 | 20.98062.8460 | 46.7963 | 41.1304 | 35.8958 | 31.1304 | 26.79633.1623 | 50.6902 | 45.1029 | 39.8843 | 35.1029 | 30.6902where dS is computed from the CEV volatility as in : dS = Volatility * sqrt( T / N ) = Volatility * sqrt( dT)What surprises me, and I would like to ask you if you experience anything similar, is that for low volatilities (15, 30, 45) the values from CEV and tree are close enough, while when the volatilities increase, the models start to diverge in the results.Is it only me (and therefore a coding error), or do you observe similar behaviour?If so, what does it mean? Since I am using CEV with q = r = 0, the differential equation govening the price of the underlying reduces to dS = sigma*dB, which should be normal... Thanks a lot in advance,gc

Excuse my ignorance, but what is CEV?If you are modelling stock prices with the usual model, but zero drift then this is STILL a lognormal model, not a gaussian one. At lower vols, the differece is less significant that at higher vols. Could this be the reason?Regards,Sam

Hi Sam, thanks for the reply, but I don't think this is the problem.The CEV (Constant Elasticity of Volatility) model assumes that the process for the stock price is:dS = (r-q)Sdt + sigma*S^alpha dW (dW usual Brownian motion)As usual r is riskfree and q dividend yield. If alpha is 1 CEV returns the usual lognormal BS results, while if alpha is 0 then the process of the underlying is:dS = (r-q)Sdt + sigma dWand therefore I believe that if r=q it is normal (hence my comparing it with the normal binomial tree).I have to study a bit more about the stability of the trees, but I am now wondering whether it could be that the tree is correct, but for large values of volatilities, it is not stable and this is the reason for the diverging values... Any ideas?Thanksgc

For general guidance, look at the late David Nelson's"Simple Binomial Processes as Diffusion Approximations".Review of Financial Studies, 3, 393-430. (1990)I suspect your particular problem is the incorrect handling ofabsorption or reflection at S=0.Regards,alan p.s. - here is the paper:http://rfs.oupjournals.org/cgi/reprint/3/3/393.pdf

Hi Alan,Thanks a lot, the article is very informative and it is exactly what I actually needed.Nelson suggest a transformation to implement binomial trees (when alpha>=1/2) that reduces number of steps and increases stability, and I will try it.More and more I am now thinking that my problem is really the stability: the prices from CEV and binomial trees start to diverge when sigma>75, that corresponds to a volatility for a stock priced 50 of 150%, so quite a high value afterall... As regards the absorption or reflection at S=0, I don't know... At the end I will be using trees for interest rate options, and I don't see any reason why, - at least in theory - the interest rate couldn't go negative (waiting for Japan to be deflationary....)Thanks a lot, your posting was very useful,gc

Hi,I have come across another idea. I read a paper about Edgeworth binomial tree and there is a really interesing idea, where edgeworth statistical series expansion is used on standartized normal distribution This solution should approximate higher moments like skewness and Kurtosis of ending distribution of continuous returns. But what about using totally diffferent ending distribution like hyperbolic one - mean = 0, variance = 1, and demanded skewness and kurtosis. We can use the same approach that was used in the paper to ensure that it will be a martingale. (read the paper - it is short) And there exist the only one binomial tree that lead to this ending distribution. I really need any advice. Am I correct ???? Thanks PeterIt is really important, because in that case I can use edgeworth binomial tree to evaluate exotic option - asian and lookback with Monte Carlo Simulation through the tree